Question

Nickel has a face-centered cubic structure and has a density of 8.90 g/cm3. what is its atomic radius?

Answer 1

density = (mass/volume) 

Mass = # atoms (in unit cell) x (mol / 6.022 x 10^23 atoms) x (58.69 g/mol) 
The last number is the atomic mass of nickel 

The number of atoms = 8(1/8) + 6(1/2) = 4 

The volume (of the entire unit cell) = side^3 = (r x 8^(1/2))^3 

6.84 g/cm3 = [4 atoms x (mol / 6.022 x 10^23 atoms) x (58.69 g/mol)] / [r x 8^(1/2)]^3 

 r = 1.36 x 10^-8 cm

Answer 2

Answer:

Atomic radius, $r=1.24\times 10^{-8}\ cm$

Explanation:

It is given that,

Density of nickel, d = 8.9 g/cm³

The density of a unit cell is given by :

$d=\dfrac{Z\times A}{N_0\times a^3}$

Where

Z = no of atoms per unit cell

A = molar mass of an element in g/mol

N₀ = Avogadro's number

a = edge length

The edge length of FCC crystal is, $a=\sqrt8r$.............(1)

r = atomic radius

Nickel has a FCC structure and for FCC, Z = 4

$a^3=\dfrac{Z\times A}{N_0\times d}$

For nickel A = 58.69 g/mol

$a^3=\dfrac{4\times 58.69}{6.022\times 10^{23}\times 8.9}$

$a^3=4.38\times 10^{-23}\ cm$

$a=3.52\times 10^{-8}\ cm$

Atomic radius is, $r=\dfrac{a}{\sqrt8}$

$r=\dfrac{3.52\times 10^{-8}}{\sqrt8}$

$r=1.24\times 10^{-8}\ cm$

Hence, this is the required solution.